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  TopK算法杂记
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  <h1 id="概述"><a href="#概述" class="headerlink" title="概述"></a>概述</h1><p>TopK的字面意思很简单，就是求给定序列的前$k$大（or小）的元素。本文由简单思路入手，对解决方案逐步发散与优化。为简化叙述，本文讨论的均为最大标准下的TopK（最小标准同理）。</p>
<span id="more"></span>
<h1 id="1-全局排序"><a href="#1-全局排序" class="headerlink" title="1. 全局排序"></a>1. 全局排序</h1><p>这是最基础的想法，也是最容易实现的。只需要将给定序列进行全局排序后，取出最大的前$k$个即为所求。这里的排序算法选择可以参考<a href="https://blog.shimmerjordan.eu.org/2022/02/25/8SortingAlgorithms/">八大排序算法笔记</a>，其最优的时间复杂度为$O(nlogn)$。</p>
<h1 id="2-局部排序"><a href="#2-局部排序" class="headerlink" title="2. 局部排序"></a>2. 局部排序</h1><p>全局排序的时间复杂度之所以高，是因为对给定序列中除TopK的元素外进行了不必要的排序。那么我们就可以考虑使用局部排序的思想来优化算法。</p>
<p>我们需要尽可能只对前$k$个最大数进行排序操作，或者只进行找到前$k$个最大数的操作。显然冒泡是一个很好的选择，因为每一趟的冒泡都可以找到当前序列的子序列最大值，也就可以保证在$k$趟冒泡的时间内找出TopK。</p>
<p><strong>代码实现：</strong></p>
<figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">def</span> <span class="title function_">bubbleTopk</span>(<span class="params">target, k</span>):</span><br><span class="line">    <span class="keyword">for</span> i <span class="keyword">in</span> <span class="built_in">range</span>(<span class="built_in">len</span>(target)):</span><br><span class="line">        <span class="keyword">if</span> i == k-<span class="number">1</span>:</span><br><span class="line">            <span class="keyword">return</span> target[-k:][::-<span class="number">1</span>]</span><br><span class="line">        <span class="keyword">for</span> j <span class="keyword">in</span> <span class="built_in">range</span>(<span class="built_in">len</span>(target)-i-<span class="number">1</span>):</span><br><span class="line">            <span class="keyword">if</span> target[j] &gt; target[j+<span class="number">1</span>]:</span><br><span class="line">                target[j], target[j+<span class="number">1</span>] = target[j+<span class="number">1</span>], target[j]</span><br><span class="line"></span><br><span class="line">target = [<span class="number">3</span>,<span class="number">1</span>,<span class="number">5</span>,<span class="number">7</span>,<span class="number">2</span>,<span class="number">4</span>,<span class="number">9</span>,<span class="number">6</span>,<span class="number">10</span>,<span class="number">8</span>]</span><br><span class="line"><span class="built_in">print</span>(bubbleTopk(target,<span class="number">4</span>))</span><br></pre></td></tr></table></figure>
<p><strong>时间复杂度：</strong>$O(nk)$</p>
<h1 id="3-堆"><a href="#3-堆" class="headerlink" title="3. 堆"></a>3. 堆</h1><p>可以发现，在上面的冒泡过程中，我们还是对TopK进行了排序，但实际上只需要取出TopK而不需要进行TopK的排序。这就引出利用堆优化的TopK算法：</p>
<p><strong>基本思路：</strong></p>
<ol>
<li><p>先用前$k$个元素生成一个小顶堆，这个小顶堆用于存储，当前最大的$k$个元素；</p>
</li>
<li><p>接着，从第$k+1$个元素开始扫描，和堆顶（堆中最小的元素）比较，如果被扫描的元素大于堆顶，则替换堆顶的元素，并调整堆，以保证堆内的$k$个元素，总是当前最大的$k$个元素；</p>
</li>
<li><p>直到扫描完所有$n-k$个元素，最终堆中的$k$个元素，就是所需的TopK。</p>
</li>
</ol>
<p><strong>代码实现：</strong></p>
<p>这里直接调用了python库<a target="_blank" rel="noopener" href="https://docs.python.org/zh-cn/3/library/heapq.html"><code>heapq</code></a>，其中<code>replace</code>方法弹出并返回 <code>heap</code>中最小的一项，同时推入新的<code>item</code>。 堆的大小不变。</p>
<figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">import</span> heapq</span><br><span class="line"><span class="keyword">def</span> <span class="title function_">heapqTopk</span>(<span class="params">target, k</span>):</span><br><span class="line">    heap = []</span><br><span class="line">    <span class="keyword">for</span> i <span class="keyword">in</span> <span class="built_in">range</span>(k):</span><br><span class="line">        heapq.heappush(heap, target[i])</span><br><span class="line">    <span class="keyword">for</span> j <span class="keyword">in</span> <span class="built_in">range</span>(k+<span class="number">1</span>,<span class="built_in">len</span>(target)):</span><br><span class="line">        <span class="keyword">if</span> target[j] &gt; heap[<span class="number">0</span>]:</span><br><span class="line">            heapq.heapreplace(heap, target[j])</span><br><span class="line">    <span class="keyword">return</span> heap</span><br></pre></td></tr></table></figure>
<p><strong>时间复杂度：</strong></p>
<p>$n$个元素扫一遍，假设运气很差，每次都入堆调整，调整时间复杂度为堆的高度，即$log(k)$，故整体时间复杂度是$O(nlogk)$。</p>
<h1 id="4-随机选择-partition"><a href="#4-随机选择-partition" class="headerlink" title="4. 随机选择+partition"></a>4. 随机选择+partition</h1><p>堆，将冒泡的TopK排序优化为了TopK不排序，节省了计算资源。堆，是求TopK的经典算法，那还有没有更快的方案呢？</p>
<p>我们知道快速排序的核心思想是<strong>分治法</strong>：</p>
<blockquote>
<p>分治法（Divide&amp;Conquer），把一个大的问题，转化为若干个子问题（Divide），每个子问题“都”解决，大的问题便随之解决（Conquer）。这里的关键词是“都”。从伪代码里可以看到，快速排序递归时，先通过partition把数组分隔为两个部分，两个部分“都”要再次递归。</p>
</blockquote>
<p>分治法有一个特例，叫<strong>减治法</strong>:</p>
<blockquote>
<p><strong>减治法</strong>（Reduce&amp;Conquer），把一个大的问题，转化为若干个子问题（Reduce），这些子问题中“<strong>只</strong>”解决一个，大的问题便随之解决（Conquer）。这里的关键词是 <strong>“只”</strong> 。</p>
</blockquote>
<p>通过分治法与减治法的描述，可以发现，分治法（快排$O(nlogn)$）的复杂度一般来说是大于减治法（二分$O(logn)$）的。我们或许可以利用利用减治法来简化分治法的终止边界条件，最终达到简化算法的目的。</p>
<p>我们知道，<strong>快排</strong>的算法核心<code>partition(arr, low, high)</code>会把序列<code>arr</code>分为两个部分（默认<code>privot = arr[low]</code>作为划分依据）：左半部分都比<code>privot</code>大，右半部分都比<code>privot</code>小。而且<code>partition</code>返回的是<code>privot</code>的最终位置<code>index</code>。</p>
<p><strong><code>partition</code>和TopK问题有什么关系呢？</strong></p>
<p>TopK是希望求出<code>arr[1,n]</code>中最大的k个数，那如果找到了<strong>第$k$大</strong>的数，做一次<code>partition</code>，不就一次性找到最大的$k$个数了么？问题变成了<code>arr[1, n]</code>中找到第$k$大的数。</p>
<p>再回过头来看看<strong>第一次</strong><code>partition</code>，划分之后：</p>
<p><code>index = partition(arr, 1, n);</code></p>
<p>如果<code>index</code>大于<code>k</code>，则说明<code>arr[index]</code>左边的元素都大于<code>k</code>，于是只递归<code>arr[1, index-1]</code>里第k大的元素即可；</p>
<p>如果<code>index</code>小于<code>k</code>，则说明说明第<code>k</code>大的元素在<code>arr[index]</code>的右边，于是只递归<code>arr[index+1, n]</code>里第<code>k-index</code>大的元素即可；</p>
<p>这就是<strong>随机选择</strong>算法Randomized Select，其<strong>代码实现</strong>如下：</p>
<figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">def</span> <span class="title function_">partition</span>(<span class="params">target, low, high</span>):</span><br><span class="line">    privot = target[low]</span><br><span class="line">    i, j = low, high</span><br><span class="line">    <span class="keyword">while</span> i &lt; j:</span><br><span class="line">        <span class="keyword">while</span> i &lt; j <span class="keyword">and</span> target[j] &lt;= privot:</span><br><span class="line">            j -= <span class="number">1</span></span><br><span class="line">        <span class="keyword">while</span> i &lt; j <span class="keyword">and</span> target[i] &gt;= privot:</span><br><span class="line">            i += <span class="number">1</span></span><br><span class="line">        target[i], target[j] = target[j], target[i]</span><br><span class="line">    target[i], target[low] = target[low], target[i]</span><br><span class="line">    <span class="keyword">return</span> i</span><br><span class="line"></span><br><span class="line"><span class="keyword">def</span> <span class="title function_">randomSelectTopk</span>(<span class="params">target, k</span>):</span><br><span class="line">    left, right = <span class="number">0</span>, <span class="built_in">len</span>(target)-<span class="number">1</span></span><br><span class="line">    index = -<span class="number">1</span></span><br><span class="line">    <span class="keyword">while</span> k != index+<span class="number">1</span>:</span><br><span class="line">        index = partition(target, left, right)</span><br><span class="line">        <span class="keyword">if</span> index &gt; k:</span><br><span class="line">            right = index - <span class="number">1</span></span><br><span class="line">        <span class="keyword">elif</span> index &lt; k:</span><br><span class="line">            left = index + <span class="number">1</span></span><br><span class="line">        <span class="keyword">else</span>:</span><br><span class="line">            <span class="keyword">return</span> target[:k]</span><br><span class="line"></span><br><span class="line">target = [<span class="number">3</span>,<span class="number">1</span>,<span class="number">5</span>,<span class="number">7</span>,<span class="number">2</span>,<span class="number">4</span>,<span class="number">9</span>,<span class="number">6</span>,<span class="number">10</span>,<span class="number">8</span>]</span><br><span class="line"><span class="built_in">print</span>(randomSelectTopk(target,<span class="number">4</span>))</span><br></pre></td></tr></table></figure>
<p><strong>时间复杂度：</strong> 这是一个典型的减治算法，递归内的两个分支，最终只会执行一个，它的时间复杂度是$O(n)$。</p>
<h1 id="Summary"><a href="#Summary" class="headerlink" title="Summary"></a>Summary</h1><p>TopK不难；其思路优化过程不简单：知其然，知其所以然。思路比结论重要。</p>
<ul>
<li><p>全局排序，$O(nlogn)$</p>
</li>
<li><p>局部排序，只排序TopK个数，$O(nk)$</p>
</li>
<li><p>堆，TopK个数也不排序了，$O(nlogk)$</p>
</li>
<li><p>随机选择+<code>partition</code>：利用分治和减治思想优化，$O(n)$</p>
</li>
</ul>
 
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